复杂度O（n^(2/3)）10^11以内的素数个数
int范围内两个相邻素数间隔不到几百
#include<cstdio>  
#include<cmath>  
using namespace std;  
#define LL long long  
const int N = 5e6 + 2;  
bool np[N];  
int prime[N], pi[N];  
int getprime()  
{  
    int cnt = 0;  
    np[0] = np[1] = true;  
    pi[0] = pi[1] = 0;  
    for(int i = 2; i < N; ++i)  
    {  
        if(!np[i]) prime[++cnt] = i;  
        pi[i] = cnt;  
        for(int j = 1; j <= cnt && i * prime[j] < N; ++j)  
        {  
            np[i * prime[j]] = true;  
            if(i % prime[j] == 0)   break;  
        }  
    }  
    return cnt;  
}  
const int M = 7;  
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;  
int phi[PM + 1][M + 1], sz[M + 1];  
void init()  
{  
    getprime();  
    sz[0] = 1;  
    for(int i = 0; i <= PM; ++i)  phi[i][0] = i;  
    for(int i = 1; i <= M; ++i)  
    {  
        sz[i] = prime[i] * sz[i - 1];  
        for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];  
    }  
}  
int sqrt2(LL x)  
{  
    LL r = (LL)sqrt(x - 0.1);  
    while(r * r <= x)   ++r;  
    return int(r - 1);  
}  
int sqrt3(LL x)  
{  
    LL r = (LL)cbrt(x - 0.1);  
    while(r * r * r <= x)   ++r;  
    return int(r - 1);  
}  
LL getphi(LL x, int s)  
{  
    if(s == 0)  return x;  
    if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];  
    if(x <= prime[s]*prime[s])   return pi[x] - s + 1;  
    if(x <= prime[s]*prime[s]*prime[s] && x < N)  
    {  
        int s2x = pi[sqrt2(x)];  
        LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;  
        for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];  
        return ans;  
    }  
    return getphi(x, s - 1) - getphi(x / prime[s], s - 1);  
}  
LL getpi(LL x)  
{  
    if(x < N)   return pi[x];  
    LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;  
    for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;  
    return ans;  
}  
LL lehmer_pi(LL x)  
{  
    if(x < N)   return pi[x];  
    int a = (int)lehmer_pi(sqrt2(sqrt2(x)));  
    int b = (int)lehmer_pi(sqrt2(x));  
    int c = (int)lehmer_pi(sqrt3(x));  
    LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2;  
    for (int i = a + 1; i <= b; i++)  
    {  
        LL w = x / prime[i];  
        sum -= lehmer_pi(w);  
        if (i > c) continue;  
        LL lim = lehmer_pi(sqrt2(w));  
        for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);  
    }  
    return sum;  
}  
int main()  
{  
    init();  
    LL n;  
    while(~scanf("%lld",&n))  
    {  
        printf("%lld\n",lehmer_pi(n));  
    }  
    return 0;  
}  


Pollard Rho 算法

typedef long long ll;
const int N = 1e5 + 7;
ll x, y, a[N];
ll max_factor;
struct BigIntegerFactor {
    const static int N = 1e6 + 7;
    ll prime[N], p[N], fac[N], sz, cnt; //多组输入注意初始化cnt = 0
    inline ll mul(ll a, ll b, ll mod) {          //WA了尝试改为__int128或慢速乘
        if (mod <= 1000000000)
            return a * b % mod;
        return (a * b - (ll)((long double)a / mod * b + 1e-8) * mod + mod) % mod;
    }
    void init(int maxn) {
        int tot = 0;
        sz = maxn - 1;
        for (int i = 1; i <= sz; ++i)
            p[i] = i;
        for (int i = 2; i <= sz; ++i) {
            if (p[i] == i)
                prime[tot++] = i;
            for (int j = 0; j < tot && 1ll * i * prime[j] <= sz; ++j) {
                p[i * prime[j]] = prime[j];
                if (i % prime[j] == 0)
                    break;
            }
        }
    }
    ll qpow(ll a, ll x, ll mod) {
        ll res = 1ll;
        while (x) {
            if (x & 1)
                res = mul(res, a, mod);
            a = mul(a, a, mod);
            x >>= 1;
        }
        return res;
    }
    bool check(ll a, ll n) {                     //二次探测原理检验n
        ll t = 0, u = n - 1;
        while (!(u & 1))
            t++, u >>= 1;
        ll x = qpow(a, u, n), xx = 0;
        while (t--) {
            xx = mul(x, x, n);
            if (xx == 1 && x != 1 && x != n - 1)
                return false;
            x = xx;
        }
        return xx == 1;
    }
    bool miller(ll n, int k) {
        if (n == 2)
            return true;
        if (n < 2 || !(n & 1))
            return false;
        if (n <= sz)
            return p[n] == n;
        for (int i = 0; i <= k; ++i) {            //测试k次
            if (!check(rand() % (n - 1) + 1, n))
                return false;
        }
        return true;
    }
    inline ll gcd(ll a, ll b) {
        return b == 0 ? a : gcd(b, a % b);
    }
    inline ll Abs(ll x) {
        return x < 0 ? -x : x;
    }
    ll Pollard_rho(ll n) {                 //基于路径倍增的Pollard_Rho算法
        ll s = 0, t = 0, c = rand() % (n - 1) + 1, v = 1, ed = 1;
        while (1) {
            for (int i = 1; i <= ed; ++i) {
                t = (mul(t, t, n) + c) % n;
                v = mul(v, Abs(t - s), n);
                if (i % 127 == 0) {
                    ll d = gcd(v, n);
                    if (d > 1)
                        return d;
                }
            }
            ll d = gcd(v, n);

            if (d > 1)
                return d;
            s = t;
            v = 1;
            ed <<= 1;
        }
    }
    void getfactor(ll n) {                         //得到所有的质因子(可能有重复的)
        if (n <= sz) {
            while (n != 1)
                fac[cnt ++ ] = p[n], n /= p[n];
            max_factor = max_factor > p[n] ? max_factor : p[n];
            return;
        }
        if (miller(n, 6)) {
            fac[cnt ++ ] = n;
            max_factor = max_factor > n ? max_factor : n;
        }
        else {
            ll d = n;
            while (d >= n)
                d = Pollard_rho(n);
            getfactor(d);
            getfactor(n / d);
        }
        return ;
    }
} Q;
int main() {
    //Q.init(N - 1);//如果代码超时且仅需要分解大数的质因数可以用这句话，否则不要用
    ll T, n;
    scanf("%lld", &T);
    while (T--) {
        max_factor = -1;
        scanf("%lld", &n);
        Q.getfactor(n);
        if(max_factor == n)
            puts("Prime");
        else printf("%lld\n", max_factor);
    }
    return 0;
}

Miller−Rabin 判定法
ll Rand() {//决定了程序的性能
    static ll x = (srand((int)time(0)), rand());
    x += 1000003;
    if (x > 1000000007)
        x -= 1000000007;
    
    return x;
}
bool Witness(ll a, ll n) {
    ll t = 0, u = n - 1;
    while (!(u & 1))
        u >>= 1, t++;
    ll x = qpow(a, u, n), y;//qpow为快速幂

    while (t--) {
        y = x * x % n;
        if (y == 1 && x != 1 && x != n - 1)
            return true;
        x = y;
    }
    return x != 1;
}
bool MillerRabin(ll n, ll s) {
    if (n == 2 || n == 3 || n == 5)
        return 1;

    if (n % 2 == 0 || n % 3 == 0 || n % 5 == 0 || n == 1)
        return 0;

    while (s--) {
        if (Witness(Rand() % (n - 1) + 1, n))
            return false;
    }
    return true;
}
    